--- title: "Crank AB 5 rad/s, piston velocity" subject: "Applied Mechanics (Dynamics)" chapter: 4 key: "semester_3/applied_mechanics_dynamics:crank-5rads-piston-velocity-coriolis" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/crank-5rads-piston-velocity-coriolis site: BCT Engineering --- # Crank AB 5 rad/s, piston velocity Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies - Asked in: 2081 Baishakh, 8+2 marks - Video: [(5/83) Angular Velocity of Piston Linkage](https://www.youtube.com/watch?v=wCo2jU70YZ4) — The Ryder Project - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/crank-5rads-piston-velocity-coriolis - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4 ## Question Crank AB rotates with a constant angular velocity of 5 rad/s. Determine the velocity of piston C and the angular velocity of link BC at the instant $\theta = 30^\circ$. Define coriolis acceleration. [Figure: crank AB = 600 mm at angle $\theta$ to the horizontal, link BC = 300 mm, piston C sliding horizontally, 150 mm offset between line of A and slider line.] ## Short answer Using relative velocity equations and the horizontal slider condition, v_C = 3.00 m/s towards A, ω_BC = 10.00 rad/s clockwise. Coriolis acceleration is the term 2ω × v_rel that arises when a point moves relative to a rotating frame; it is perpendicular to the relative velocity. ## Answer Assume the crank $AB$ turns counter-clockwise at $\omega_{AB}=5$ rad/s, with $A$ as origin, $B$ above and to the right of $A$, and the slider line $0.15$ m above $A$ with $C$ to the right of $B$ (the magnitudes of the answers are the same for the other sense of rotation). #### Geometry $$ \vec r_{B/A} = 0.6(\cos30^\circ\hat i + \sin30^\circ\hat j) = (0.5196\,\hat i + 0.3000\,\hat j)\ \text{m} $$ $B$ is $0.300 - 0.150 = 0.150$ m above the slider line, so $\sin\phi = 0.150/0.300 = 0.5$ and $\phi=30^\circ$ (link $BC$ slopes down to the right): $$ \vec r_{C/B} = 0.3(\cos30^\circ\hat i - \sin30^\circ\hat j) = (0.2598\,\hat i - 0.1500\,\hat j)\ \text{m} $$ #### Velocities $$ \vec v_B = \omega_{AB}\hat k\times\vec r_{B/A} = 5\hat k\times(0.5196\hat i + 0.3000\hat j) = (-1.500\,\hat i + 2.598\,\hat j)\ \text{m/s} $$ $$ \vec v_C = \vec v_B + \omega_{BC}\hat k\times\vec r_{C/B} = \vec v_B + \omega_{BC}(0.1500\,\hat i + 0.2598\,\hat j) $$ The slider moves horizontally, so the $\hat j$ component of $\vec v_C$ is zero: $$ 0 = 2.598 + 0.2598\,\omega_{BC} \;\Rightarrow\; \omega_{BC} = -10.00\ \text{rad/s}\ \text{(clockwise)} $$ $$ v_C = -1.500 + 0.1500(-10.00) = -3.00\ \text{m/s} $$ **Answer: $v_C = 3.00$ m/s (horizontal, towards $A$), $\omega_{BC} = 10.00$ rad/s clockwise.** #### Coriolis acceleration **Coriolis acceleration** is the extra acceleration a point has when it moves relative to a rotating (moving) frame or body. If a point $P$ moves with relative velocity $\vec v_{rel}$ on a body rotating with angular velocity $\vec\omega$, then $$ \vec a_P = \vec a_{P'} + \vec a_{rel} + 2\vec\omega\times\vec v_{rel} $$ where $\vec a_{P'}$ is the acceleration of the coincident point of the rotating body and $\vec a_{rel}$ is the acceleration of $P$ relative to the body. The term $\vec a_c = 2\vec\omega\times\vec v_{rel}$ is the Coriolis acceleration. Its magnitude is $2\omega v_{rel}$ and its direction is that of $\vec v_{rel}$ rotated through $90^\circ$ in the sense of $\omega$. It exists only when the point slides on the rotating body (for example a collar sliding on a rotating arm, or a slider in a slotted rotating link); it is zero for two points fixed in the same rigid body, such as the crank pin and piston pin of a connecting rod. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the angle of link BC found? A. B is 0.150 m above the slider line, so sin(phi) = 0.150/0.300 = 0.5 and phi = 30 deg B. phi equals the crank angle only by assumption of 60 deg C. tan(phi) = 0.6/0.3 D. phi is 0 because C is horizontal Answer: A. The vertical height of B above the slider line is 0.300 - 0.150 = 0.150 m, and BC = 0.3 m. (Key point: Link angle) 2. What condition determines w_BC? A. The i component of vC is zero B. The j component of vC is zero, since the slider moves horizontally C. vB is zero D. w_BC = w_AB Answer: B. 0 = 2.598 + 0.2598 w_BC gives w_BC = -10 rad/s (10 rad/s clockwise). (Key point: Slider constraint) 3. What is the speed of piston C? A. 1.50 m/s B. 3.00 m/s C. 3.00 m/s vertical D. 5.20 m/s Answer: B. vC = -1.500 + 0.1500(-10) = -3.00 m/s, i.e. 3.00 m/s horizontal towards A. (Key point: Piston velocity) 4. Coriolis acceleration is defined as A. the acceleration w^2 r of a point on a rotating body B. the extra term 2 w x v_rel for a point moving relative to a rotating body C. the tangential acceleration alpha r D. the acceleration of the centre of mass Answer: B. It appears when a point moves relative to a rotating frame, with magnitude 2 w v_rel. 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